# Proof of Double Power Law of Logarithms

## Formula

$\log_{b^y}{m^x}$ $\,=\,$ $\Big(\dfrac{x}{y}\Big)\log_{b}{m}$

The double power rule of logarithms is a mathematical identity which is used to find the value of logarithm of a quantity by expressing quantity and base quantity of logarithmic term in exponential notation.

### Proof

$p$ and $q$ are two quantities and assume they are expressed in exponential form as $m^{\displaystyle x}$ and $n^{\displaystyle y}$ respectively.

$p \,=\, m^{\displaystyle x}$ and $q \,=\, n^{\displaystyle y}$

The value of logarithm of $p$ to $q$ is written as $\log_{q}{p}$ in mathematics. Actually, $p \,=\, m^{\displaystyle x}$ and $q \,=\, n^{\displaystyle y}$.

Therefore, $\log_{q}{p}$ $\,=\,$ $\log_{n^y}{m^{\displaystyle x}}$

#### Find Log of quantity in exponential form

Take $t \,=\, b^y$ and the logarithmic function can be written as follows.

$\implies \log_{b^y}{m^{\displaystyle x}}$ $\,=\,$ $\log_{t}{m^{\displaystyle x}}$

According to Power law of Logarithms, the log of a quantity in exponential form to a base is equal to the product of exponent and log of the base of exponential term to same base.

$\implies \log_{b^y}{m^{\displaystyle x}}$ $\,=\,$ $x\log_{t}{m}$

Now, replace the actual value of the base $t$.

$\,\,\, \therefore \,\,\,\,\,\, \log_{b^y}{m^{\displaystyle x}}$ $\,=\,$ $x\log_{b^y}{m}$

#### Find Log of quantity to base in exponential form

It is time to find the value of log of $m$ to a base which is expressed in exponential form as $b^{y}$. It can be done by the base power rule of logarithm.

$\,\,\, \therefore \,\,\,\,\,\, \log_{b^y}{m}$ $\,=\,$ $\Big(\dfrac{1}{y}\Big)\log_{b}{m}$

#### Combine results of both steps

Now, recollect the results of above two steps once.

$(1) \,\,\,\,\,\,$ $\log_{b^y}{m^{\displaystyle x}}$ $\,=\,$ $x\log_{b^y}{m}$

$(2) \,\,\,\,\,\,$ $\log_{b^y}{m}$ $\,=\,$ $\Big(\dfrac{1}{y}\Big)\log_{b}{m}$

Now, combine both results to a log property to find the value of log of a quantity in exponential form to base in exponential form.

$\implies \log_{b^y}{m^x}$ $\,=\,$ $x \times \log_{b^y}{m}$

$\implies \log_{b^y}{m^x}$ $\,=\,$ $x \times \Big(\dfrac{1}{y}\Big)\log_{b}{m}$

$\,\,\, \therefore \,\,\,\,\,\, \log_{b^y}{m^x}$ $\,=\,$ $\Big(\dfrac{x}{y}\Big)\log_{b}{m}$

Thus, the double power rule of logarithms is derived in algebraic form and it can be used as an identity in mathematics.

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Jun 26, 2023

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